Question

Cinnabar eyes (cn) and reduced bristles (rd) are autosomal recessive characters in Drosophila. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 males were then crossed to the F1 females to obtain the F2. Of the 400 F2 offspring obtained, 292 were wild type, 9 were cinnabar, 7 were reduced, and 92 were reduced and cinnabar.

Briefly explain (2 sentences) these results and estimate the distance between the cn and rd loci.

Answer 1

Answer: The homozygous wild type female genotype = cn+rd+/cn+rd+

The genotype for reduced, cinnabar male =cnrd/cnrd

The cross will give rise to F1 genotype = cn+cn/rd+rd

The F1 female will produce the gamates cn+rd+, cn+rd, cnrd+ and cnrd

While the F1 male will produce the gamates cn+rd+ and cnrd

NOTE: In drosophila the male does not undergo recombination during meiosis and hence the above gamates will be formed in the male.

The F2 generation results are not in accordance with Mendel’s dihybrid cross (9:3:3:1) therefore the genes must be linked.

Now in F2 generation of the 400 offsprings

292 are wild type means having the phenotype as cn+rd+/cn+rd+(parental) or cn+rd+/cnrd (parental)   or cnrd/cn+rd+(parental) or cn+rd/cn+rd+(recombinant) or cnrd+/cn+rd+ ( recombinant).

The 9 cinnabar and 7 reduced are recombinants moreover in above 292 , the last two genotypes are showing recombination.

Assuming that these recombinants are in the same proportion i.e 9 and 7

Recombination frequency = no of recombinants/ total offsprings multiplied by 100

It will be equal to 2*(7+9)/400 multiplied by 100

The answer is 8%

Therefore the distance between cn and rd genes is 8cM